What is the solution for y^'=xy^{1/2},y(0)= 1/16 x^4 ?

The solution for y^'=xy^{1/2},y(0)= 1/16 x^4 is y=(x^4}{16}+\frac{c_{1}x^2)/4+(c_{1}^2)/4

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y^'=xy^{1/2},y(0)= 1/16 x^4

y^{^{'}} = x y^{\frac{1}{2}}, y (0) = \frac{1}{16} x^{4}

y = \frac{x ^{4}}{16} + \frac{c _{1} x ^{2}}{4} + \frac{c _{1}^{2}}{4}

Solution steps

y^{'} (x) = x y^{\frac{1}{2}}

Solve separable ODE:

y = \frac{x ^{4}}{16} + \frac{c _{1} x ^{2}}{4} + \frac{c _{1}^{2}}{4}

y = \frac{x ^{4}}{16} + \frac{c _{1} x ^{2}}{4} + \frac{c _{1}^{2}}{4}

e^{x} (\frac{d y}{d x}) = 1 + y^{2}

\frac{d y}{d x} = 4 x 36 - y^{2}

\frac{d y}{d t} = \frac{- t}{y}, y = 4

y = sin (2) (ln (x^{2} - 3 x + 2)) y^{^{'}}

\frac{d v}{d t} = 90 - \frac{4}{100} v, v = 0, t = 0

What is the solution for y^'=xy^{1/2},y(0)= 1/16 x^4 ?
The solution for y^'=xy^{1/2},y(0)= 1/16 x^4 is y=(x^4}{16}+\frac{c_{1}x^2)/4+(c_{1}^2)/4